The pressure exerted by an electromagnetic wave of intensity `I (Wm^(-2))` on a non-reflecting surface is ( c is the velocity of light)

Options

A. `Ic`
B. `Ic^(2)`
C. `I//c`
D. `I//c^(2)`

Explanation

To find the pressure exerted by an electromagnetic wave of intensity \( I \) on a non-reflecting surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definition of Intensity**:
The intensity \( I \) of an electromagnetic wave is defined as the power \( P \) per unit area \( A \):
\[ I = \frac{P}{A} \]

2. **Relate Power to Work**:
Power can also be expressed in terms of work done over time. The formula for power is:
\[ P = \frac{\text{Work}}{\text{Time}} = \frac{W}{t} \]

3. **Express Work in Terms of Force and Displacement**:
Work done \( W \) can be expressed as the product of force \( F \) and displacement \( S \):
\[ W = F \cdot S \]

4. **Substitute Displacement with Speed**:
Since displacement over time gives us speed, we can relate displacement to speed \( C \) (the speed of light):
\[ S = C \cdot t \]
Therefore, substituting this into the work equation gives:
\[ W = F \cdot (C \cdot t) \]

5. **Combine the Equations**:
Now, substituting the expression for work back into the power equation:
\[ P = \frac{F \cdot (C \cdot t)}{t} = F \cdot C \]

6. **Relate Force to Pressure**:
Pressure \( P \) is defined as force per unit area:
\[ P = \frac{F}{A} \]
Therefore, we can express force in terms of pressure:
\[ F = P \cdot A \]

7. **Substitute Force Back into the Power Equation**:
Substituting \( F \) into the power equation gives:
\[ I = \frac{F \cdot C}{A} = \frac{(P \cdot A) \cdot C}{A} = P \cdot C \]

8. **Solve for Pressure**:
Rearranging the equation to solve for pressure \( P \):
\[ P = \frac{I}{C} \]

### Final Answer:
Thus, the pressure exerted by an electromagnetic wave of intensity \( I \) on a non-reflecting surface is:
\[ P = \frac{I}{C} \]